What is the integral of (-4x^3)/(sqrt(x^2+9)) ?

The integral of (-4x^3)/(sqrt(x^2+9)) is -4(1/3 (x^2+9)^{3/2}-9sqrt(x^2+9))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (-4x^3)/(sqrt(x^2+9))

\int \frac{- 4 x ^{3}}{x ^{2} + 9} d x

- 4 (\frac{1}{3} (x^{2} + 9)^{\frac{3}{2}} - 9 x^{2} + 9) + C

Solution steps

\int \frac{- 4 x ^{3}}{x ^{2} + 9} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - 4 \cdot \int \frac{x ^{3}}{x ^{2} + 9} d x

Apply u-substitution

: \int u^{2} - 9 d u

= - 4 \cdot \int u^{2} - 9 d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - 4 (\int u^{2} d u - \int 9 d u)

\int u^{2} d u = \frac{u ^{3}}{3}

\int 9 d u = 9 u

= - 4 (\frac{u ^{3}}{3} - 9 u)

Substitute back

u = x^{2} + 9

= - 4 (\frac{( x ^{2} + 9 ) ^{3}}{3} - 9 x^{2} + 9)

Simplify

- 4 (\frac{( x ^{2} + 9 ) ^{3}}{3} - 9 x^{2} + 9) : - 4 (\frac{1}{3} (x^{2} + 9)^{\frac{3}{2}} - 9 x^{2} + 9)

= - 4 (\frac{1}{3} (x^{2} + 9)^{\frac{3}{2}} - 9 x^{2} + 9)

Add a constant to the solution

= - 4 (\frac{1}{3} (x^{2} + 9)^{\frac{3}{2}} - 9 x^{2} + 9) + C

\frac{d}{d x} (ln (1 + 3 x))

\int 12 x^{3} - 4 x d x

(1 + t) (2 t^{2} - 5)

f (x) = (\frac{4 x ^{2}}{2 - x})^{3}

\frac{11 - 3}{p - 1}

What is the integral of (-4x^3)/(sqrt(x^2+9)) ?
The integral of (-4x^3)/(sqrt(x^2+9)) is -4(1/3 (x^2+9)^{3/2}-9sqrt(x^2+9))+C